I want to show that for any square matrix $A \in \mathbb{C}^{n \times n}$ and any unitary $X \in \mathbb{C}^{n\times n}$ (that is $XX^* = X^* X = I$) the following holds true

$$ \text{Re} \{\text{trace}(X^* A)\} \leq \sum_{i =1}^{rank(A)}\sigma_{i}(A) \,, $$

where $\text{Re}\{\cdot\}$ is "a real part of", and $\sigma_i(A)$ are the singular values of matrix $A$.

I was able to show the result in what I feel is an awkward way. I am looking for a shorter way which it is based on the well established properties of singular value decomposition and trace, without any lengthy matrix decompositions as I did.

Here is my solution:

(Note: $A = U \Sigma V^*$ - is the SVD of the matrix $A$) $$ \text{Re}\{tr(X^*A)\} = \text{Re}\{tr(X^*U\Sigma V^*)\} \underbrace{=}_{\text{tr(AB) = tr(BA)}} \text{Re}\{tr(V^*XU\Sigma ) \} =\text{Re}\{tr(S\Sigma)\}\,, $$

where $S= V^*XU$ is unitary matrix, thus the rows$\backslash$columns are orthonormal (in particular columns have norm $1$).

Writing $S = \widetilde{S_1} + \cdots + \widetilde{S_n}$, where the matrices $\widetilde{S_i}\in \mathbb{C}^{n \times n}$ are matrices where all the elements are zero except for the column $i$ which equals to column $i$ in $S$ (that is I decompose $S$ to matrices containing single column of $S$).

This yields

$$ S\Sigma = \sigma_1 \widetilde{S_1} + \cdots + \sigma_n \widetilde{S_n}\,. $$

Therefor I get

$$ Re\{tr(S\Sigma)\} = Re\{tr(\sigma_1 \widetilde{S_1})+ \cdots + tr(\sigma_n \widetilde{S_n})\} = Re\{\sigma_1tr(\widetilde{S_1})+ \cdots + \sigma_n tr( \widetilde{S_n})\}\,. $$ Now, in general every element in the vector $v_i$ can be bounded by the norm of the vector, since columns of the matrix $S$ are of norm one, the elements in $\widetilde{S}_i$ can be bounded by $1$ to yield $$ |tr(\widetilde{S}_i)| \leq 1\, ,\forall i\,.$$

Overall this implies

$$ Re\{tr(S\Sigma)\} = Re\{\sigma_1tr(\widetilde{S_1})+ \cdots + \sigma_n tr( \widetilde{S_n})\}\leq \sum_{i= 1}^{n} \sigma_i = \sum_{i= 1}^{rank(A)} \sigma_i $$

Added later: Here I found another question which proves the same result.


The polar decomposition of $A$ is $A = W R$ where $W$ is unitary and $R = (A^* A)^{1/2}$ has eigenvalues $\sigma_j$, the singular values of $A$ (including zeros). In terms of the SVD, you could write this as $A = U \Sigma V^* = (U V^*)(V \Sigma V^*)$. If $v_j$ are the corresponding orthonormal basis of eigenvectors of $R$,

$$ \text{tr}(X^* A) = \text{tr}(X^*W R) = \sum_j v_j^* X^* W R v_j = \sum_j \sigma_j v_j^* X^*W v_j$$ Now use the fact that $|v_j^* X^*W v_j| \le \|v_j\| \|X^* W v_j\| = 1$.

  • $\begingroup$ I realized from your answer that for any set of orthonormal basis $\{v_i\}_{i=1}^n$ and any square matrix A, we can write $tr(A) = \sum_{i=1}^n tr(v_i^* Av_i)$ - that would simplify my argument. $\endgroup$ – them Aug 20 '16 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.